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Velocity required to increase friction?

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1.2 g pebble is lodged in the tread of a 0.76-m automobile tire, held in place by static friction that can be at most 3.6 N. The car starts from rest and gradually accelerates on a straight road. How fast is the car moving when the pebble flies out of the tire tread?


    2. Relevant equations
    I looked at the Chapter references and found v=ωR or v=2∏R/T where T=the period of the circle.


    3.Attempts
    1.I thought of Torque of the wheel but I can't find the Inertia (mass of tire not given) and I have no idea what the magnitude of the force rolling the tire.

    2.I tried a=F/m and got a=3.6N/.0012kg=3000 m/s2.

    3.I thought of the torque as a result of the static friction and computed τ=3.6N*0.38m=1.368N*m but how do I apply that to find the velocity of the car? I can't use τ=Iα because while I have the mass of the pebble I don't have the mass of the tire and if I used the pebble I don't know the radius of the pebble (if there is one). So this method won't work.

    4.V=2∏R/T is useless because the problem does not give me time at all. And I would need time to find the period and/or velocity.

    My largest obstacle so far in this course is my trouble in decomposing the components. Would anyone mind helping me get started in solving this problem?

    Thanks
     
  2. jcsd
  3. Mar 6, 2012 #2
    what forces are acting on something that is moving in a circle?
     
  4. Mar 6, 2012 #3

    gneill

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    Staff: Mentor

    Hi aJaguarinRed, Welcome to Physics Forums.

    You have a particle of some small mass m moving in a circle of some radius r. Right away you should recognize circular motion, and think about what forces or accelerations are involved. You should be able to find a rotation rate, ω, for the wheel which will create a net force that will dislodge the particle (Hint: where around the circumference is that force or acceleration going to be largest? What accelerations are contributing?).

    You should also be able to find a relationship between the car's speed and the rotation rate of the wheel. In time t the car moves a distance d*t. How many 'circumferences' of the wheel is that?
     
  5. Mar 6, 2012 #4
    Thanks, but what has got me baffled is how can I find the velocity or acceleration without the time??? I tried F=mv2/r →3.6N=((.0012kg)v2)/0.38m and got the answer 33.76 m/s which is wrong according to the back of the book (answer given is 17 m/s). Okay, I found an explanation on (sadly) Yahoo! Answers http://answers.yahoo.com/question/index?qid=20100408125401AAt1hQg and well, wow. I actually solved the problem minus one step which was to divide the v by 2 (33.76/2=16.8≈17 m/s). Can anyone explain to me why you need to divide by two?

    Thanks again for your hint. I'm pleased that with your hints I figured a way to do it but not pleased that I'm not fully understanding the answer.
     
  6. Mar 6, 2012 #5

    gneill

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    Staff: Mentor

    The center of the wheel is moving along with the car. The bottom of the wheel where it touches the road is momentarily at rest with respect to the road. So in the frame of reference of the road, there's no centripetal force on the stone at that point. The top of the wheel, on the other hand, is moving at twice the car's speed (it has to go faster that the car does in order to "get in front" of the center of the wheel). The tangential speed there is thus twice the car's speed. So whatever tangential speed you found that will dislodge the stone, the car will be doing half that.
     
  7. Mar 6, 2012 #6

    gneill

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    Staff: Mentor

    The above was an argument that I remembered seeing elsewhere. Now that I think about it I believe it to be specious. The wheel is essentially stationary in an inertial frame of reference along with the car. As such, the relative velocity of its components with respect to another inertial frame (in this case the the one in which the road is stationary) will not affect the physics in its own frame.

    A simple though experiment will confirm this. Suppose you have wheel mounted on a bench in a lab, and the wheel is free to rotate about a horizontal axis. If you calculate the required angular velocity to dislodge the stone you will use ω2r for the centripetal acceleration (and take into account gravity also affecting the stone w.r.t. rotational position). The angular velocity in turn gives you the tangential speed of the wheel.

    Now let's say that you just found out that your lab was sitting in the back of a truck moving at a constant velocity v along a straight level road. If the wheel is oriented correctly, it will behave the same as the wheels of the car, with its bottom most point momentarily being stationary w.r.t. the road, and its top moving at 2v w.r.t. the road. Yet the physics of the rotating wheel cannot change simply by making this observation. So the "2v" argument fails.

    That said, your own result of about 33.8m/s is closer to being correct. I believe the book's answer is wrong.

    The only modifications I would make to your solution would be to include the influence of gravity on the stone, so the affect of centripetal acceleration will depend upon where in the wheel's revolution it is (the weight of the stone adds to the friction force holding it in place when it's at the top, and acts against the friction force at the bottom).
     
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